Question: $f(x, y) = 9 - 3x^3y - 3xy^3$ What are all the critical points of $f$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $(0, 0)$ (Choice B) B $(1, 3)$ (Choice C) C $(3, 1)$ (Choice D) D There are no critical points.
Solution: A critical point of a scalar field $f$ is where $\nabla f = \bold{0}$. [What's that bolded 0?] Let's find the gradient of $f$ ! $\nabla f = \begin{bmatrix} -9x^2y - 3y^3 \\ \\ -3x^3 - 9xy^2 \end{bmatrix}$ We want each component of the gradient to equal zero, so we want to solve the system of equations below. $\begin{cases} -9x^2y - 3y^3 = 0 \\ \\ -3x^3 - 9xy^2 = 0 \end{cases}$ We can simplify the system of equations: $\begin{cases} y(3x^2 + y^2) = 0 \\ \\ x(x^2 + 3y^2) = 0 \end{cases}$ If $x = 0$, then the first equation becomes $y^3 = 0$, so $y = 0$. If $y = 0$, then the second equation becomes $x^3 = 0$, so $x = 0$. There is no solution in which $x$ and $y$ are both nonzero, because the terms in parentheses are always positive, so neither equation could equal zero. There is only the solution at $(0, 0)$. Therefore, $f$ has a critical point at $(0, 0)$.